We're so Banach Take AI Quiz by dokuDoku Math 🔍 ### Prerequisites - [No Holes Allowed! What's a Complete Space?](https://www.noteblogdoku.com/blog/no-holes-allowed-whats-a-complete-space) - [Closed, Open, Clopen](https://www.noteblogdoku.com/blog/closed-open-clopen) - [Metric Space vs Inner Product Space](https://www.noteblogdoku.com/blog/metric-space-vs-inner-product-space) ## Prelude to What's a Banach Space? Formally: **A normed vector space is a Banach space if it is complete with respect to the metric induced by the norm.** To discuss this we are going to start from what a normed space is. Then we will discuss what benefits we can get from having the space be complete with respect to the norm. The definition of a vector space can be seen in the **Appendix**. To motivate Banach spaces, consider that for finite dimensional spaces, such as $\mathbb{R}^n$, Cauchy sequences converge. Many problems however in differential equations, Fourier analysis, optimization, and quantum mechanics live in infinite dimensional spaces. In these spaces convergence can fail if completeness is not guaranteed. Banach spaces are great to work with since they guarantee that the limits are within the space. Completeness may appear to be merely a technical condition, however many theorems and behaviors we take for granted in $\mathbb{R}^n$ for example rely on this property. ## What's a Normed Space? A normed space is a vector space $V$ with a norm, $||\cdot ||$ assigning non-negative lengths to each vector. When we take the norm of a vector, we are taking the length of said vector, which is the distance that the vector is from the origin of the vector space. Properties of the norm: 1. Positive Definiteness: $||x|| \geq 0$ and $||x|| = 0$ if and only if $x=0$ 2. Absolute Homogeneity: $||\alpha x|| = |\alpha|\ ||x||$ where $\alpha$ is a scalar member of a field - Typically the field is either the reals $\mathbb{R}$ or the complex numbers $\mathbb{C}$ 3. Triangle Inequality: $||x+y|| \leq ||x|| + ||y||$ Note that since the norm induces a metric, $d(x,y) = ||x-y||$, that a normed space is automatically a metric space. Note that for a metric space that there are properties that the distance function must fulfill which are: Properties of Metric Space's Distance Function 1. Non-negative distance $d(x,y) \geq 0$ 2. Identity of Indiscernibles $d(x,y) = 0$ if and only if $x=y$ 3. Symmetry: $d(x,y) = d(y,x)$ 4. Triangle Inequality: $d(x,z) \leq d(x,y) + d(y,z)$ We can see that the distance function's properties of non-negative distance and identity of discernibles are easily fulfilled by $d(x,y) = ||x-y||$. Additionally, we can see the symmetry condition is fulfilled with some short algebra: First take $(y-x) = -(x-y)$. Then we are going to show that $||y-x|| = ||x-y||$ $$ (y-x) = -(x-y) \rightarrow ||y-x|| = |-1|\ ||x-y|| = ||x-y||$$ Thus giving $$ ||y-x|| = ||x-y|| $$ Similarly, we can show the triangle inequality for the normed space $$ d(x,z) = ||x-z||$$ $$ ||x-z|| = ||(x-y) + (y-z)|| \leq ||x-y|| + ||y-z|| $$ Giving us $$ d(x,z) \leq d(x,y) + d(y,z) $$ Great! We have shown that all normed spaces are metric spaces! We can see that normed spaces give us access to be able to talk about - Distances between elements - Convergence and limits of sequences ## Why would we want a *complete* normed space? The definition of completeness of a metric space is: **Every Cauchy Sequence is a Convergent Sequence** As explained in [No Holes Allowed! What's a Complete Space?](https://www.noteblogdoku.com/blog/no-holes-allowed-whats-a-complete-space) a complete space is defined such that for a given metric space, there are no holes. Completeness guarantees that the limits of sequences exist within the metric space. So when we have that for all possible sequences in the metric space, that they approach only elements in the set in the metric space defined by $(X,d)$. That's great but if we complete our normed space, i.e. make into a Banach space, *what benefits do we get over just an ordinary normed space?* The only difference between a complete and incomplete normed space, is that for a complete normed space, **limits are guaranteed to exist within the space**. From this we get several benefits. 1. **Banach Fixed-Point Theorem (Contraction Mapping Theorem)** If $X$ is complete and $T: X \rightarrow X$ is a contraction, then $T$ has a unique fixed point, and the iterates $x_{n+1} = Tx_n$ converge to it from any starting point. **Note:** A contraction is a mapping such that: $$d(T(x), T(y)) \leq k \cdot d(x,y)$$ for some fixed $0 \leq k <1$, $\forall x,y \in X$. This is basically saying that a contraction is a map that reduces the distance between $x,y \in X$ to a fraction (factor $k$) of its original. Repeated application, $n$ times, gives that $$ d(T^n(x), T^n(y)) \leq k^n d(x,y)$$ Since $k^n \rightarrow 0$, we get that $d(T^n(x), T^n(y)) \rightarrow 0$ as $n \rightarrow \infty$. The theorem states that there exists a unique fixed point $x^* \in X$ such that $$ T(x^*) = x^* $$ Furthermore, for any starting point $x_0 \in X$ the sequence defined by $$ x_{n+1} = T(x_n) $$ converges to $x^*$, and the completeness of $X$ guarantees the limit exists in $X$. This gives us the existence and uniqueness of a fixed point. Many problems can be reformulated as fixed point problems, allowing Banach's theorem to establish existence and uniqueness of solutions. Some of these problemes include ODE's, PDE's, integral equations, and optimization problems. The following strategy is typically utilized for these types of problems. 1. Construct a function space $X$ 2. Define a map $T$ 3. Show $T$ is a contraction 4. Apply Banach's theorem This is very useful. 2. **Absolute-Convergence Criterion** A normed space is complete if and only if every absolutely convergent series converges to an element of the space. To define an **absolutely convergent series** let $\{v_n\}^\infty_{n=1}$ be a sequence of vectors in $V$. Then the series $\sum_{n=1}^\infty v_n$ is **absolutely summable** if the series $$ \sum_{n=1}^\infty ||v_n||$$ converges in $\mathbb{R}$. Equivalently, the sequence of partial sums $$ \{ \sum_{n=1}^m ||v_n||\}_{m=1}^\infty $$ converges in $\mathbb{R}$. Note that a series $\sum_{n=1}^\infty v_n$ is said to **converge** if its sequence of partial sums $$s_m = \sum_{n=1}^m v_n$$ converges to some element $s \in V$. The **Absolute-Convergence Criterion** states that $V$ is complete if and only if every absolutely summable series in $V$ converges to an element of the space. It should be stated that the Absolute Convergence Criterion is actually a requirement for Banach spaces: **$V$ is a Banach space if and only if every absolutely summable series in $V$ converges to an element in $V$.** The Absolute-Convergence Criterion provides one of the most useful characterizations of Banach spaces. It allows completeness to be verified through the behavior of infinite series rather than directly through Cauchy sequences. Since many objects in analysis are constructed as limits of series, the criterion guarantees that absolutely summable series converge to elements already contained in the space, ensuring that the space is closed under these fundamental limiting processes. Engineers care about this theorem since this states that infinite approximations converge to a valid element within the vector space. Examples include Fourier Series for signal processing and control theory, where the response of a system can be represented as a series expansion. Banach spaces are also useful for engineers since many proofs follow this pattern: 1. Construct sequence of approximations 2. Show the increments become small $$ \sum ||v_n|| < \infty $$ 3. Invoke Absolute-Convergence Criterion within the Banach space 4. Conclude approximations converge to a valid solution. The theorem can be used to state that if the total accumulated error is finite, then an infinite process converges within the space. ## Examples of Banach Spaces There are three common examples of Banach spaces including 1. $\mathbb{R}^n$ 2. $\mathbb{C}^n$ 3. $C[a,b]$ with supremum norm $||f|| = sup|f(t)|$ - Continuous functions on the interval $[a,b] \in \mathbb{R}$ ### $\mathbb{R}^n$ This is the standard space of all $n$ dimensional vectors $$\mathbb{R}^n = \{(x_1,x_2,...,x_n): x_i \in \mathbb{R}\}$$ We are going to assume the Eucledian norm here $$||x||_2 = \sqrt{x_1^2 + ... x_n^2}$$ You may be asking why is this space Banach? Note that a Cauchy Sequence in $\mathbb{R}^n$ $$x^{(1)},x^{(2)},...$$ is really $n$ coordinate wise Cauchy sequences in $\mathbb{R}^n$ (one for each dimension). Thus $$x^{(n)} \rightarrow (x_1,...,x_n) \in \mathbb{R}^n$$ So $\mathbb{R}^n$ is complete. Thus the space $$(\mathbb{R}^n, ||\cdot ||)$$ is a Banach space for this norm. ### $\mathbb{C}^n$ This is basically the same as $\mathbb{R}^n$, except every coordinate is complex. $$\mathbb{C}^n = \{(x_1,x_2,...,x_n): x_i \in \mathbb{C}\}$$ ### $C[a,b]$ with supremum norm $||f|| = sup|f(t)|$ The formal definition of this space is $$ C[a,b] = \{f:[a,b] \rightarrow \mathbb{R} | \text{ f continuous}\}$$ Some examples of these functions (when restricted to $x \in [a,b]$) are - $f(x) = x$ - $g(x) = sin(x)$ - $h(x) = e^x $ Firstly, this space has vector space operations including Addition $$ (f+g)(x) = f(x) + g(x) $$ Scalar Multiplication $$(cf)(x) = c \cdot f(x) $$ Proof is left to the reader for these two properties :) The norm we defined on $C[a,b]$ is the supremum norm $$ ||f||_\infty = sup_{x \in [a,b]} |f(x)| $$ which is the least upper bound (largest) element in the function between $[a,b]$. Because $f$ is continuous on a closed interval, this supremum exists. But why is this space $C[a,b]$ Banach? Suppose that the sequence of continuous functions $f_1, f_2, ...$ is Cauchy in the sup norm, meaning that as we progress through the sequence the distance between the terms become arbitrarily close, with respect to the supremum norm. We can see an illustration of this in **Figure 1**. <img src="/media/images/20260608_043142_06_07_2026_21_22_07_cauchy_sequence_pairwise_distance.gif" alt="06 07 2026 21 22 07 cauchy sequence pairwise distance" style="max-width: 100%; height: auto;"> <center> $\textbf{Figure 1}$: Cauchy sequence in $C([a,b],\mathbb{R})$ with last term $f_n$ on top and previous term $f_{n-1}$ on bottom and their difference below. Supremum-metric distance $d(f_n,f_{n-1})=\sup_x|f_n-f_{n-1}| \rightarrow 0$ as $n \rightarrow \infty$ </center> Recall the definition of a **Cauchy Sequence** A sequence $(x_n)$ in a metric space $(X,d)$ is a Cauchy sequence if for every $\epsilon > 0\ \exists N$ such that $$ d(x_m,x_n) < \epsilon \quad \forall m,n \geq N $$ Likewise, for our Cauchy sequence $f_1, f_2, ...$ we have that within this defined metric space with the supremum norm, this Cauchy sequence means that $$ d(f_m, f_n) < \epsilon \quad \forall m,n \geq N$$ i.e. $$ sup_x |f_n(x) - f_m(x)| < \epsilon \quad \forall m,n \geq N $$ Which is basically stating that $$ sup_x |f_n(x) - f_m(x)| \rightarrow 0 $$ From $$ \sup_{x\in[a,b]} |f_n(x)-f_m(x)| < \epsilon \qquad \forall m,n\ge N, $$ we know that $(f_n)$ is Cauchy in the supremum norm. Now fix any point $x\in[a,b]$. Since $$ |f_n(x)-f_m(x)| \le \sup_{t\in[a,b]} |f_n(t)-f_m(t)|, $$ it follows that $$ |f_n(x)-f_m(x)|<\epsilon \qquad \forall m,n\ge N. $$ Thus, for each fixed $x$, the sequence of real numbers $$ f_1(x),f_2(x),\ldots $$ is a Cauchy sequence in $\mathbb{R}$. Since $\mathbb{R}$ is complete, there exists a number $f(x)\in\mathbb{R}$ such that $$ f_n(x)\to f(x). $$ This defines a function $$ f:[a,b]\to\mathbb{R}. $$ ### Showing $f_n \to f$ uniformly We now show convergence in the supremum norm. Let $\epsilon>0$. Since $(f_n)$ is Cauchy, there exists $N$ such that $$ \sup_x |f_n(x)-f_m(x)| < \epsilon \qquad \forall m,n\ge N. $$ Fix $n\ge N$. Taking the limit as $m\to\infty$, and using the fact that $f_m(x)\to f(x)$ pointwise, we obtain $$ |f_n(x)-f(x)| = \lim_{m\to\infty}|f_n(x)-f_m(x)| \le \epsilon $$ for every $x\in[a,b]$. Therefore $$ \sup_x |f_n(x)-f(x)| \le \epsilon, $$ which means $$ \|f_n-f\|_\infty \to 0. $$ Hence $f_n$ converges to $f$ in the supremum norm. ### Showing $f$ is continuous Each $f_n$ is continuous, and we have just shown that $$ f_n \to f $$ uniformly on $[a,b]$. A standard theorem from analysis states: *The uniform limit of continuous functions is continuous.* Therefore $f$ is continuous on $[a,b]$, so $$ f \in C[a,b]. $$ ### Conclusion We started with an arbitrary Cauchy sequence $(f_n)$ in $(C[a,b],\|\cdot\|_\infty)$. Using the completeness of $\mathbb{R}$, we constructed a function $f:[a,b]\to\mathbb{R}$ such that $$ f_n(x)\to f(x) \qquad \text{for every } x\in[a,b]. $$ We then showed that this convergence is actually uniform: $$ \|f_n-f\|_\infty \to 0. $$ Since the uniform limit of continuous functions is continuous, we have $$ f \in C[a,b]. $$ Thus every Cauchy sequence in $(C[a,b],\|\cdot\|_\infty)$ converges to an element of $C[a,b]$. Therefore, $$ (C[a,b],\|\cdot\|_\infty) \text{ is a Banach space.} $$ --- ## Appendix ### If a sequence is convergent then every subsequence is too We are going to prove: $$\text{If $(a_n)$ converges to $L$, then every subsequence $(a_{n_k})$ also converges to $L$}$$ Note that a subsequence is determined by a strictly increasing sequence of indices $$ n_1 < n_2 < ...$$ and $(a_{n_k})_{k \geq 1}$ is the resulting sequence of terms. We are going to state a small lemma (without proof) that will be useful: $$ \text{For a strictly increasing sequence of natural numbers } n_1 < n_2 < ... \text{ we have } n_k \geq k \text{ for every } k$$ To prove the original theorem, assume $a_n \rightarrow L$. We want to show $a_{n_k} \rightarrow L$ as $k \rightarrow \infty$. Let $\epsilon > 0$ be arbitrary. By the definition of convergence of $(a_n)$, there exists an $N \in \mathbb{N}$ such that $$ |a_n - L| < \epsilon \quad \forall n \geq N$$ Take any $k \geq N$. By the lemma $n_k \geq k \geq N$, since $n_k \geq N$, the inequality applies with $n = n_k$ giving $$ |a_{n_k} - L| < \epsilon $$ Thus for every $\epsilon > 0$ we found an index $N$ such that $|a_{n_k} -L| < \epsilon$ when $k \geq N$. This is exactly the statement $$ lim_{k \rightarrow \infty} a_{n_k} = L $$ ### Vector Space Definition Let $ \mathbb{F} $ be a field (most commonly $ \mathbb{F} = \mathbb{R} $ or $ \mathbb{F} = \mathbb{C} $). A vector space over $ \mathbb{F} $ is a set $ V $ equipped with two operations: Vector addition: $ +: V \times V \to V $ Scalar multiplication: $ \cdot: \mathbb{F} \times V \to V $ such that the following 10 axioms hold for all vectors $ \mathbf{u}, \mathbf{v}, \mathbf{w} \in V $ and all scalars $ \alpha, \beta \in \mathbb{F} $: Axioms for Vector Addition Associativity: $ (\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w}) $ Commutativity: $ \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} $ Identity element: There exists a zero vector $ \mathbf{0} \in V $ such that $ \mathbf{u} + \mathbf{0} = \mathbf{u} $ Inverse element: For every $ \mathbf{u} \in V $, there exists $ -\mathbf{u} \in V $ such that $ \mathbf{u} + (-\mathbf{u}) = \mathbf{0} $ Axioms for Scalar Multiplication Distributivity (scalar over vectors): $ \alpha(\mathbf{u} + \mathbf{v}) = \alpha\mathbf{u} + \alpha\mathbf{v} $ Distributivity (vectors over scalars): $ (\alpha + \beta)\mathbf{u} = \alpha\mathbf{u} + \beta\mathbf{u} $ Compatibility: $ \alpha(\beta\mathbf{u}) = (\alpha\beta)\mathbf{u} $ Scalar identity: $1 \cdot \mathbf{u} = \mathbf{u}$ Additional Axioms Closure under addition: $ \mathbf{u} + \mathbf{v} \in V $ Closure under scalar multiplication: $ \alpha\mathbf{u} \in V $ ### Prerequisites - [No Holes Allowed! What's a Complete Space?](https://www.noteblogdoku.com/blog/no-holes-allowed-whats-a-complete-space) - [Closed, Open, Clopen](https://www.noteblogdoku.com/blog/closed-open-clopen) - [Metric Space vs Inner Product Space](https://www.noteblogdoku.com/blog/metric-space-vs-inner-product-space) ## Prelude to What's a Banach Space? Formally: **A normed vector space is a Banach space if it is complete with respect to the metric induced by the norm.** To discuss this we are going to start from what a normed space is. Then we will discuss what benefits we can get from having the space be complete with respect to the norm. The definition of a vector space can be seen in the **Appendix**. To motivate Banach spaces, consider that for finite dimensional spaces, such as $\mathbb{R}^n$, Cauchy sequences converge. Many problems however in differential equations, Fourier analysis, optimization, and quantum mechanics live in infinite dimensional spaces. In these spaces convergence can fail if completeness is not guaranteed. Banach spaces are great to work with since they guarantee that the limits are within the space. Completeness may appear to be merely a technical condition, however many theorems and behaviors we take for granted in $\mathbb{R}^n$ for example rely on this property. ## What's a Normed Space? A normed space is a vector space $V$ with a norm, $||\cdot ||$ assigning non-negative lengths to each vector. When we take the norm of a vector, we are taking the length of said vector, which is the distance that the vector is from the origin of the vector space. Properties of the norm: 1. Positive Definiteness: $||x|| \geq 0$ and $||x|| = 0$ if and only if $x=0$ 2. Absolute Homogeneity: $||\alpha x|| = |\alpha|\ ||x||$ where $\alpha$ is a scalar member of a field - Typically the field is either the reals $\mathbb{R}$ or the complex numbers $\mathbb{C}$ 3. Triangle Inequality: $||x+y|| \leq ||x|| + ||y||$ Note that since the norm induces a metric, $d(x,y) = ||x-y||$, that a normed space is automatically a metric space. Note that for a metric space that there are properties that the distance function must fulfill which are: Properties of Metric Space's Distance Function 1. Non-negative distance $d(x,y) \geq 0$ 2. Identity of Indiscernibles $d(x,y) = 0$ if and only if $x=y$ 3. Symmetry: $d(x,y) = d(y,x)$ 4. Triangle Inequality: $d(x,z) \leq d(x,y) + d(y,z)$ We can see that the distance function's properties of non-negative distance and identity of discernibles are easily fulfilled by $d(x,y) = ||x-y||$. Additionally, we can see the symmetry condition is fulfilled with some short algebra: First take $(y-x) = -(x-y)$. Then we are going to show that $||y-x|| = ||x-y||$ $$ (y-x) = -(x-y) \rightarrow ||y-x|| = |-1|\ ||x-y|| = ||x-y||$$ Thus giving $$ ||y-x|| = ||x-y|| $$ Similarly, we can show the triangle inequality for the normed space $$ d(x,z) = ||x-z||$$ $$ ||x-z|| = ||(x-y) + (y-z)|| \leq ||x-y|| + ||y-z|| $$ Giving us $$ d(x,z) \leq d(x,y) + d(y,z) $$ Great! We have shown that all normed spaces are metric spaces! We can see that normed spaces give us access to be able to talk about - Distances between elements - Convergence and limits of sequences ## Why would we want a *complete* normed space? The definition of completeness of a metric space is: **Every Cauchy Sequence is a Convergent Sequence** As explained in [No Holes Allowed! What's a Complete Space?](https://www.noteblogdoku.com/blog/no-holes-allowed-whats-a-complete-space) a complete space is defined such that for a given metric space, there are no holes. Completeness guarantees that the limits of sequences exist within the metric space. So when we have that for all possible sequences in the metric space, that they approach only elements in the set in the metric space defined by $(X,d)$. That's great but if we complete our normed space, i.e. make into a Banach space, *what benefits do we get over just an ordinary normed space?* The only difference between a complete and incomplete normed space, is that for a complete normed space, **limits are guaranteed to exist within the space**. From this we get several benefits. 1. **Banach Fixed-Point Theorem (Contraction Mapping Theorem)** If $X$ is complete and $T: X \rightarrow X$ is a contraction, then $T$ has a unique fixed point, and the iterates $x_{n+1} = Tx_n$ converge to it from any starting point. **Note:** A contraction is a mapping such that: $$d(T(x), T(y)) \leq k \cdot d(x,y)$$ for some fixed $0 \leq k <1$, $\forall x,y \in X$. This is basically saying that a contraction is a map that reduces the distance between $x,y \in X$ to a fraction (factor $k$) of its original. Repeated application, $n$ times, gives that $$ d(T^n(x), T^n(y)) \leq k^n d(x,y)$$ Since $k^n \rightarrow 0$, we get that $d(T^n(x), T^n(y)) \rightarrow 0$ as $n \rightarrow \infty$. The theorem states that there exists a unique fixed point $x^* \in X$ such that $$ T(x^*) = x^* $$ Furthermore, for any starting point $x_0 \in X$ the sequence defined by $$ x_{n+1} = T(x_n) $$ converges to $x^*$, and the completeness of $X$ guarantees the limit exists in $X$. This gives us the existence and uniqueness of a fixed point. Many problems can be reformulated as fixed point problems, allowing Banach's theorem to establish existence and uniqueness of solutions. Some of these problemes include ODE's, PDE's, integral equations, and optimization problems. The following strategy is typically utilized for these types of problems. 1. Construct a function space $X$ 2. Define a map $T$ 3. Show $T$ is a contraction 4. Apply Banach's theorem This is very useful. 2. **Absolute-Convergence Criterion** A normed space is complete if and only if every absolutely convergent series converges to an element of the space. To define an **absolutely convergent series** let $\{v_n\}^\infty_{n=1}$ be a sequence of vectors in $V$. Then the series $\sum_{n=1}^\infty v_n$ is **absolutely summable** if the series $$ \sum_{n=1}^\infty ||v_n||$$ converges in $\mathbb{R}$. Equivalently, the sequence of partial sums $$ \{ \sum_{n=1}^m ||v_n||\}_{m=1}^\infty $$ converges in $\mathbb{R}$. Note that a series $\sum_{n=1}^\infty v_n$ is said to **converge** if its sequence of partial sums $$s_m = \sum_{n=1}^m v_n$$ converges to some element $s \in V$. The **Absolute-Convergence Criterion** states that $V$ is complete if and only if every absolutely summable series in $V$ converges to an element of the space. It should be stated that the Absolute Convergence Criterion is actually a requirement for Banach spaces: **$V$ is a Banach space if and only if every absolutely summable series in $V$ converges to an element in $V$.** The Absolute-Convergence Criterion provides one of the most useful characterizations of Banach spaces. It allows completeness to be verified through the behavior of infinite series rather than directly through Cauchy sequences. Since many objects in analysis are constructed as limits of series, the criterion guarantees that absolutely summable series converge to elements already contained in the space, ensuring that the space is closed under these fundamental limiting processes. Engineers care about this theorem since this states that infinite approximations converge to a valid element within the vector space. Examples include Fourier Series for signal processing and control theory, where the response of a system can be represented as a series expansion. Banach spaces are also useful for engineers since many proofs follow this pattern: 1. Construct sequence of approximations 2. Show the increments become small $$ \sum ||v_n|| < \infty $$ 3. Invoke Absolute-Convergence Criterion within the Banach space 4. Conclude approximations converge to a valid solution. The theorem can be used to state that if the total accumulated error is finite, then an infinite process converges within the space. ## Examples of Banach Spaces There are three common examples of Banach spaces including 1. $\mathbb{R}^n$ 2. $\mathbb{C}^n$ 3. $C[a,b]$ with supremum norm $||f|| = sup|f(t)|$ - Continuous functions on the interval $[a,b] \in \mathbb{R}$ ### $\mathbb{R}^n$ This is the standard space of all $n$ dimensional vectors $$\mathbb{R}^n = \{(x_1,x_2,...,x_n): x_i \in \mathbb{R}\}$$ We are going to assume the Eucledian norm here $$||x||_2 = \sqrt{x_1^2 + ... x_n^2}$$ You may be asking why is this space Banach? Note that a Cauchy Sequence in $\mathbb{R}^n$ $$x^{(1)},x^{(2)},...$$ is really $n$ coordinate wise Cauchy sequences in $\mathbb{R}^n$ (one for each dimension). Thus $$x^{(n)} \rightarrow (x_1,...,x_n) \in \mathbb{R}^n$$ So $\mathbb{R}^n$ is complete. Thus the space $$(\mathbb{R}^n, ||\cdot ||)$$ is a Banach space for this norm. ### $\mathbb{C}^n$ This is basically the same as $\mathbb{R}^n$, except every coordinate is complex. $$\mathbb{C}^n = \{(x_1,x_2,...,x_n): x_i \in \mathbb{C}\}$$ ### $C[a,b]$ with supremum norm $||f|| = sup|f(t)|$ The formal definition of this space is $$ C[a,b] = \{f:[a,b] \rightarrow \mathbb{R} | \text{ f continuous}\}$$ Some examples of these functions (when restricted to $x \in [a,b]$) are - $f(x) = x$ - $g(x) = sin(x)$ - $h(x) = e^x $ Firstly, this space has vector space operations including Addition $$ (f+g)(x) = f(x) + g(x) $$ Scalar Multiplication $$(cf)(x) = c \cdot f(x) $$ Proof is left to the reader for these two properties :) The norm we defined on $C[a,b]$ is the supremum norm $$ ||f||_\infty = sup_{x \in [a,b]} |f(x)| $$ which is the least upper bound (largest) element in the function between $[a,b]$. Because $f$ is continuous on a closed interval, this supremum exists. But why is this space $C[a,b]$ Banach? Suppose that the sequence of continuous functions $f_1, f_2, ...$ is Cauchy in the sup norm, meaning that as we progress through the sequence the distance between the terms become arbitrarily close, with respect to the supremum norm. We can see an illustration of this in **Figure 1**. <img src="/media/images/20260608_043142_06_07_2026_21_22_07_cauchy_sequence_pairwise_distance.gif" alt="06 07 2026 21 22 07 cauchy sequence pairwise distance" style="max-width: 100%; height: auto;"> <center> $\textbf{Figure 1}$: Cauchy sequence in $C([a,b],\mathbb{R})$ with last term $f_n$ on top and previous term $f_{n-1}$ on bottom and their difference below. Supremum-metric distance $d(f_n,f_{n-1})=\sup_x|f_n-f_{n-1}| \rightarrow 0$ as $n \rightarrow \infty$ </center> Recall the definition of a **Cauchy Sequence** A sequence $(x_n)$ in a metric space $(X,d)$ is a Cauchy sequence if for every $\epsilon > 0\ \exists N$ such that $$ d(x_m,x_n) < \epsilon \quad \forall m,n \geq N $$ Likewise, for our Cauchy sequence $f_1, f_2, ...$ we have that within this defined metric space with the supremum norm, this Cauchy sequence means that $$ d(f_m, f_n) < \epsilon \quad \forall m,n \geq N$$ i.e. $$ sup_x |f_n(x) - f_m(x)| < \epsilon \quad \forall m,n \geq N $$ Which is basically stating that $$ sup_x |f_n(x) - f_m(x)| \rightarrow 0 $$ From $$ \sup_{x\in[a,b]} |f_n(x)-f_m(x)| < \epsilon \qquad \forall m,n\ge N, $$ we know that $(f_n)$ is Cauchy in the supremum norm. Now fix any point $x\in[a,b]$. Since $$ |f_n(x)-f_m(x)| \le \sup_{t\in[a,b]} |f_n(t)-f_m(t)|, $$ it follows that $$ |f_n(x)-f_m(x)|<\epsilon \qquad \forall m,n\ge N. $$ Thus, for each fixed $x$, the sequence of real numbers $$ f_1(x),f_2(x),\ldots $$ is a Cauchy sequence in $\mathbb{R}$. Since $\mathbb{R}$ is complete, there exists a number $f(x)\in\mathbb{R}$ such that $$ f_n(x)\to f(x). $$ This defines a function $$ f:[a,b]\to\mathbb{R}. $$ ### Showing $f_n \to f$ uniformly We now show convergence in the supremum norm. Let $\epsilon>0$. Since $(f_n)$ is Cauchy, there exists $N$ such that $$ \sup_x |f_n(x)-f_m(x)| < \epsilon \qquad \forall m,n\ge N. $$ Fix $n\ge N$. Taking the limit as $m\to\infty$, and using the fact that $f_m(x)\to f(x)$ pointwise, we obtain $$ |f_n(x)-f(x)| = \lim_{m\to\infty}|f_n(x)-f_m(x)| \le \epsilon $$ for every $x\in[a,b]$. Therefore $$ \sup_x |f_n(x)-f(x)| \le \epsilon, $$ which means $$ \|f_n-f\|_\infty \to 0. $$ Hence $f_n$ converges to $f$ in the supremum norm. ### Showing $f$ is continuous Each $f_n$ is continuous, and we have just shown that $$ f_n \to f $$ uniformly on $[a,b]$. A standard theorem from analysis states: *The uniform limit of continuous functions is continuous.* Therefore $f$ is continuous on $[a,b]$, so $$ f \in C[a,b]. $$ ### Conclusion We started with an arbitrary Cauchy sequence $(f_n)$ in $(C[a,b],\|\cdot\|_\infty)$. Using the completeness of $\mathbb{R}$, we constructed a function $f:[a,b]\to\mathbb{R}$ such that $$ f_n(x)\to f(x) \qquad \text{for every } x\in[a,b]. $$ We then showed that this convergence is actually uniform: $$ \|f_n-f\|_\infty \to 0. $$ Since the uniform limit of continuous functions is continuous, we have $$ f \in C[a,b]. $$ Thus every Cauchy sequence in $(C[a,b],\|\cdot\|_\infty)$ converges to an element of $C[a,b]$. Therefore, $$ (C[a,b],\|\cdot\|_\infty) \text{ is a Banach space.} $$ --- ## Appendix ### If a sequence is convergent then every subsequence is too We are going to prove: $$\text{If $(a_n)$ converges to $L$, then every subsequence $(a_{n_k})$ also converges to $L$}$$ Note that a subsequence is determined by a strictly increasing sequence of indices $$ n_1 < n_2 < ...$$ and $(a_{n_k})_{k \geq 1}$ is the resulting sequence of terms. We are going to state a small lemma (without proof) that will be useful: $$ \text{For a strictly increasing sequence of natural numbers } n_1 < n_2 < ... \text{ we have } n_k \geq k \text{ for every } k$$ To prove the original theorem, assume $a_n \rightarrow L$. We want to show $a_{n_k} \rightarrow L$ as $k \rightarrow \infty$. Let $\epsilon > 0$ be arbitrary. By the definition of convergence of $(a_n)$, there exists an $N \in \mathbb{N}$ such that $$ |a_n - L| < \epsilon \quad \forall n \geq N$$ Take any $k \geq N$. By the lemma $n_k \geq k \geq N$, since $n_k \geq N$, the inequality applies with $n = n_k$ giving $$ |a_{n_k} - L| < \epsilon $$ Thus for every $\epsilon > 0$ we found an index $N$ such that $|a_{n_k} -L| < \epsilon$ when $k \geq N$. This is exactly the statement $$ lim_{k \rightarrow \infty} a_{n_k} = L $$ ### Vector Space Definition Let $ \mathbb{F} $ be a field (most commonly $ \mathbb{F} = \mathbb{R} $ or $ \mathbb{F} = \mathbb{C} $). A vector space over $ \mathbb{F} $ is a set $ V $ equipped with two operations: Vector addition: $ +: V \times V \to V $ Scalar multiplication: $ \cdot: \mathbb{F} \times V \to V $ such that the following 10 axioms hold for all vectors $ \mathbf{u}, \mathbf{v}, \mathbf{w} \in V $ and all scalars $ \alpha, \beta \in \mathbb{F} $: Axioms for Vector Addition Associativity: $ (\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w}) $ Commutativity: $ \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} $ Identity element: There exists a zero vector $ \mathbf{0} \in V $ such that $ \mathbf{u} + \mathbf{0} = \mathbf{u} $ Inverse element: For every $ \mathbf{u} \in V $, there exists $ -\mathbf{u} \in V $ such that $ \mathbf{u} + (-\mathbf{u}) = \mathbf{0} $ Axioms for Scalar Multiplication Distributivity (scalar over vectors): $ \alpha(\mathbf{u} + \mathbf{v}) = \alpha\mathbf{u} + \alpha\mathbf{v} $ Distributivity (vectors over scalars): $ (\alpha + \beta)\mathbf{u} = \alpha\mathbf{u} + \beta\mathbf{u} $ Compatibility: $ \alpha(\beta\mathbf{u}) = (\alpha\beta)\mathbf{u} $ Scalar identity: $1 \cdot \mathbf{u} = \mathbf{u}$ Additional Axioms Closure under addition: $ \mathbf{u} + \mathbf{v} \in V $ Closure under scalar multiplication: $ \alpha\mathbf{u} \in V $ Comments (0) Please log in to comment. 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